Question

Evaluate the difinite integral: ${\int }_0^1(x{e}^x+\sin \frac{\pi x}{4})dx$

Answer 1

${\int }_0^1(x{e}^x+\sin \frac{\pi x}{4})dx$
Let, $F\left(x\right)=\int (x{e}^x+\sin \frac{\pi x}{4})dx$
$\stackrel{\underset{}{\int x{e}^{x}dx}dx}{I_1}+\stackrel{\int \sin \left(\frac{\pi x}{4}\right)}{I_2}$

For $I_1$:
$\int x{e}^{x}dx$
Using integration by parts
$=x\int e^{x}dx-\int [\left(\frac{dx}{dx}\right)\int e^{x}dx]dx$
$=x{e}^x-\int e^{x}dx$
$=x{e}^x-e^x$
$=e^x(x-1)$

For $I_2$ :
$\int \sin \left(\frac{\pi x}{4}\right)dx$
$=\frac{1}{\frac{\pi }{4}}(-\cos \left(\frac{\pi x}{4}\right))$
$=\frac{-4}{\pi }\cos \left(\frac{\pi x}{4}\right)$
Putting $I_1$ and $I_2$
$F\left(x\right)=\int x{e}^{x}dx+\int \sin \frac{\pi }{4}xdx$
$=e^x(x-1)-\frac{4}{\pi }\cos \left(\frac{\pi x}{4}\right)$
Now,
${\int }_0^1(x{e}^x+\sin \frac{\pi x}{4})dx$
$=F\left(1\right)-F\left(0\right)$
$=\left(e^1\right(1-1)-\frac{\pi }{4}\cos \left(\frac{\pi \times 1}{4}\right))-\left(e^0\right(0-1)+\frac{4}{\pi }\cos \left(\frac{\pi \times 0}{4}\right))$
$=0-\frac{4}{\pi }\cos \frac{\pi }{4}+1+\frac{4}{\pi }\cos 0$
$=-\frac{4}{\pi \sqrt{2}}+1+\frac{4}{\pi }$
$=1+\frac{4}{\pi }(1-\frac{1}{\sqrt{2}})$
$=1+\frac{4-2\sqrt{2}}{\pi }$